Two sides of a plot measure 32 m and 24 m a | vedclass

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(D) Let the plot be $ABCD$ with $AB = 32 \ m$,$BC = 24 \ m$,and $\angle B = 90^{\circ}$. The other two sides are $AD = 25 \ m$ and $CD = 25 \ m$.First

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(D) Let the plot be $ABCD$ with $AB = 32 \ m$,$BC = 24 \ m$,and $\angle B = 90^{\circ}$. The other two sides are $AD = 25 \ m$ and $CD = 25 \ m$.First,calculate the diagonal $AC$ using the Pythagorean theorem in $	riangle ABC$:$AC = \sqrt{AB^{2} + BC^{2}} = \sqrt{32^{2} + 24^{2}} = \sqrt{1024 + 576} = \sqrt{1600} = 40 \ m$.Now,calculate the area of $	riangle ABC$:Area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 24 	imes 32 = 384 \ m^{2}$.Next,calculate the area of $	riangle ADC$ with sides $40 \ m$,$25 \ m$,and $25 \ m$ using Heron's formula.Semi-perimeter $s = \frac{40 + 25 + 25}{2} = \frac{90}{2} = 45 \ m$.Area of $	riangle ADC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{45(45-40)(45-25)(45-25)} = \sqrt{45 	imes 5 	imes 20 	imes 20} = \sqrt{225 	imes 400} = 15 	imes 20 = 300 \ m^{2}$.Total area of the plot = Area of $	riangle ABC$ + Area of $	riangle ADC = 384 + 300 = 684 \ m^{2}$.

Two sides of a plot measure $32 \ m$ and $24 \ m$ and the angle between them is a perfect right angle. The other two sides measure $25 \ m$ each. The area of the plot in $m^{2}$ is:

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